Question

find the ratio T7/8:T1/2 for first order reaction.

Answer 1

here's is the correct answer

Answer 2

The ratio of $\frac{t_{\frac{7}{8}}}{t_{\frac{1}{2}}}$ for first order reaction is 3: 1

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

a)  $t_{\frac{7}{8}}$ of reaction  

$t=\frac{2.303}{k}\log\frac{a}{a-\frac{7}{8}}$

$t=\frac{2.303}{k}\log\frac{a}{\frac{a}{8}}$

$t=\frac{2.303}{k}\times 0.90$

b) $t_{\frac{1}{2}}$ of reaction  

$t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{a}{a-\frac{a}{2}}$

$t_{\frac{1}{2}}=\frac{2.303}{k}\times 0.30$

$\frac{t_{\frac{7}{8}}}{t_{\frac{1}{2}}}=\frac{\frac{2.303}{k}\times 0.90}{\frac{2.303}{k}\times 0.30}$

$\frac{t_{\frac{7}{8}}}{t_{\frac{1}{2}}}=3$

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